Easy INTEGER +4 / -1 PYQ · JEE Mains 2024

Number of compounds from the following with zero dipole moment is _________.

$$\mathrm{HF}, \mathrm{H}_2, \mathrm{H}_2 \mathrm{~S}, \mathrm{CO}_2, \mathrm{NH}_3, \mathrm{BF}_3, \mathrm{CH}_4, \mathrm{CHCl}_3, \mathrm{SiF}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{BeF}_2$$

Answer (integer) 6

Solution

<p>A molecule has <strong>zero dipole moment</strong> when its <strong>bond dipoles cancel by symmetry</strong> (or when bonds are nonpolar).</p> <table> <thead> <tr> <th>Species</th> <th>Shape (VSEPR)</th> <th>Dipole moment?</th> <th>Reason</th> </tr> </thead> <tbody> <tr> <td>HF</td> <td>linear</td> <td>nonzero</td> <td>polar bond, no cancellation</td> </tr> <tr> <td>$\mathrm{H}_2$</td> <td>linear</td> <td><strong>zero</strong></td> <td>homonuclear, nonpolar</td> </tr> <tr> <td>$\mathrm{H}_2\mathrm{S}$</td> <td>bent</td> <td>nonzero</td> <td>bent shape → net dipole</td> </tr> <tr> <td>$\mathrm{CO}_2$</td> <td>linear</td> <td><strong>zero</strong></td> <td>symmetric linear cancellation</td> </tr> <tr> <td>$\mathrm{NH}_3$</td> <td>trigonal pyramidal</td> <td>nonzero</td> <td>lone pair breaks symmetry</td> </tr> <tr> <td>$\mathrm{BF}_3$</td> <td>trigonal planar</td> <td><strong>zero</strong></td> <td>symmetric planar cancellation</td> </tr> <tr> <td>$\mathrm{CH}_4$</td> <td>tetrahedral</td> <td><strong>zero</strong></td> <td>symmetric tetrahedral cancellation</td> </tr> <tr> <td>$\mathrm{CHCl}_3$</td> <td>tetrahedral</td> <td>nonzero</td> <td>not symmetric ($3\ \mathrm{Cl}$ and $1\ \mathrm{H}$)</td> </tr> <tr> <td>$\mathrm{SiF}_4$</td> <td>tetrahedral</td> <td><strong>zero</strong></td> <td>symmetric tetrahedral cancellation</td> </tr> <tr> <td>$\mathrm{H}_2\mathrm{O}$</td> <td>bent</td> <td>nonzero</td> <td>bent shape → net dipole</td> </tr> <tr> <td>$\mathrm{BeF}_2$</td> <td>linear</td> <td><strong>zero</strong></td> <td>symmetric linear cancellation</td> </tr> </tbody> </table> <p>Compounds with <strong>zero dipole moment</strong>:<br>$\mathrm{H}_2,\ \mathrm{CO}_2,\ \mathrm{BF}_3,\ \mathrm{CH}_4,\ \mathrm{SiF}_4,\ \mathrm{BeF}_2$</p> <p><strong>Number = $6$.</strong></p>

About this question

Subject: Chemistry · Chapter: Chemical Bonding and Molecular Structure · Topic: Ionic and Covalent Bonding

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